combining and sorting two arrays Java?

Question

So basically there are two separate presorted arrays, and you have to combine them and sort them (without sort() methods of course). Here is my code:

public static void main(String[] args) {

    int a [] = {3,5,7,9,12,14, 15};
    int b [] = {6 ,7, 10};
    int j = 0;

    //output array should be 3,5,6,7,7,9,10,12,14,15

    int c [] = new int[a.length+b.length];//10 values

    for (int i = 0;i<b.length;i++){
        while(b[i]>a[j]){
            c[j] = a[j] ;
            j++;    
         }

        if(b[i] == a[j]){
            c[j] = b[i];
            c[j+1] = a[j];
        }

        c[j] = b[i];
        j++;
    }

    for(int i = 0;i<c.length;i++)
        System.out.println(c[i]);
    }

I'm guessing the zeros I am getting are from a mistake in one of the booleans (< & >), but I cant seem to figure it out. It works fine for the first 4, but once I get to the repeating 7's, it just goes crazy.

Please help me understand, don't just change the code.

Solution 1

This is how it should be in a simple way:

public static void main(String[] args) {

    int a [] = {3,5,7,9,12,14, 15};
    int b [] = {6 ,7, 10};
    int j = 0, k = 0;

    //output array should be 3,5,6,7,7,9,10,12,14,15

    int c [] = new int[a.length+b.length];//10 values

    // we're filling c with the next appropriate number
    // we start with checking a[0] and b[0] till we add
    // all the elements
    for (int i = 0; i < c.length; i++) {
        // if both "a" and "b" have elements left to check
        if (j < a.length && k < b.length) {
            // check if "b" has a smaller element
            if (b[k] < a[j]) {
                // if so add it to "c"
                c[i] = b[k];
                k++;
            }
            // if "a" has a smaller element
            else {
                // add it to "c"
                c[i] = a[j];
                j++;
            }       
        }
        // if there are no more elements to check in "a"
        // but there are still elements to check in "b"
        else if (k < b.length) {
            // add those elements in "b" to "c"
            c[i] = b[k];
            k++;
        }
        // if there are no more elements to check in "b"
        // but there are still elements to check in "a"
        else {
            // add those elements in "a" to "c"
            c[i] = a[j];
            j++;
        }
    }

    for(int i = 0; i < c.length; i++)
        System.out.println(c[i]);
}

Hope it helps.

Solution 2

You can try this code.

public static void main(String[] args) {
    int a[] = { 3, 5, 7, 9, 12, 14, 15 };
    int b[] = { 6, 7, 10 };

    // output array should be 3,5,6,7,7,9,10,12,14,15

    int alen = a.length;
    int blen = b.length;
    int c[] = new int[a.length + b.length];// 10 values

    int s[] = null;
    int[] l = null;

    if (alen < blen) {
        s = a;
        l = b;
    } else {
        s = b;
        l = a;
    }
            // Constructing Combined Array
    for (int i = 0, p = 0; i < c.length; i++, p++) {
        if (i == s.length) {
            p = 0;
        }
        if (i < s.length) {
            c[i] = s[p];
        } else {
            c[i] = l[p];
        }
    }
            //Sorting the C array 
    for (int i = 1; i < c.length; i++) {
        int j = i;
        int B = c[i];
        while ((j > 0) && (c[j - 1] > B)) {
            c[j] = c[j - 1];
            j--;
        }
        c[j] = B;
    }

    for (int i = 0; i < c.length; i++)
        System.out.print(c[i]);
}

Comments

[peacemaker]

Simple links should be comments, not answers

[pb2q]

please provide some substantive content in your answers, not just links to related content

[ArtemStorozhuk]

@peacemaker copied and pasted algorithm from the article. Is it OK now? :)

[Leigh]

Did you see their comment: Please help me understand, don't just change the code ? Can you elaborate on what is wrong with their current code?

[ilalex]

why are you need int s[] = null; int[] l = null;?

[K_Anas]

the problem is that he is pointing the same variable j for the cell index of tab a and tab c

[Akhi]

s[] is for storing the smaller array and l[] is for the larger array.if we know which of the 2 is smaller we can construct c(combined) easily.